Lemma 4.11 from Varadhan's Probability Theory $X$ is a countable space with each singleton measurable. Let $\pi(x,\cdot)$ be a transition probability giving a time-homogeneous Markov process.
We also have the powers
$$
\pi^{(n)}(a,b):= \sum_c \pi(c,b)\pi^{(n-1)}(a,c)
$$
which roughly represent the probability of going from $a$ to $b$ in $n$ steps.
We pick an initial point $x\in X$ and use the transition matrix to obtain a probability space $(X^{\mathbb{N}\cup\{0\}}, P_x)$.
We have Lemma 4.11 from the book/notes
Lemma 4.11 If for a (not necessarily irreducible) chain starting from $x$, the probability of ever visiting $y$ is positive, then so is the probability of visiting $y$ before returning to $x$.
Proof.  Assume that for the chain starting from $x$ the probability of visiting
$y$ before returning to $x$ is zero. But when it returns to $x$ it starts afresh and so will not visit $y$ until it returns again. This reasoning can be repeated and so the chain
will have to visit $x$ infinitely often before visiting $y$. But this will use up all the
time and so it cannot visit $y$ at all.
Question 1: What does the italicized text mean (mathematically)? I realize that
$$
P_x(\{x\}\times\{x_1\}\times \dots\times\{x_n\}\times X \times \dots) = \pi(x,x_1)\dots\pi(x_{n-1},x_n)
$$
and that
$$P_x(X\times \dots \times \{x\}\times \{x_{n+1}\}\times\dots \times \{x_{n+k}\}\times X\times \dots ) = \pi^{(n)}(x,x)\pi(x,x_{n+1})\dots \pi(x_{n+k-1}, x_{n+k}).$$
Question 2: From the italicized text and the above observation, are we implicitly making the assumption that $\pi^{(n)}(x,x)>0$ for some $n$? This seems to come from irreducibility which we don't have.
 A: Suppose we start the process from $x$. Let $\tau_i$ be the $i$th return time to $x$, and $\sigma$ be the first time such that $y$ is hit. For convenience, define $\tau_0 = 0$. I'll assume that $y \neq x$.
The lemma is saying that if $P(\sigma < \infty) > 0$, then it also follows that $P(\sigma < \tau_1) > 0.$

For the proof, the idea is that the $\tau_n$ form a partition of the time axis (up to degeneracy induced by them possibly becoming infinite after a point).But then \begin{align} P(\sigma < \infty) &= \sum_{n \ge 1} P(\tau_{n-1} < \sigma < \tau_n) \\ &=\sum_{n \ge 1} P(\sigma < \tau_n|\sigma > \tau_{n-1}) P(\sigma > \tau_{n-1})\\ & = P(\sigma < \tau_1) \sum_{n \ge 1} P(\sigma > \tau_{n-1}),\end{align} where we've used the Markov property (and the fact that the process is at $x$ at each of the times $\tau_n$) to argue $$P(\sigma < \tau_n | \sigma> \tau_{n-1}) = P(\sigma < \tau_1 | \sigma > \tau_0) = P(\sigma < \tau_1).$$ So, if $P(\sigma < \tau_1) = 0$, then $P(\sigma < \infty) = 0$, contradicting the assumption.

This doesn't require the chain to be irreducible (or even $x$ to be recurrent). In fact, if $P(\tau_1 < \infty) = 0,$ then the result is trivial, since $P(\sigma < \tau_1) = P(\sigma < \infty)$ in this case.
Also, in my opinion, the proof as written in the question is strongly preferable - all this notation is not really needed to convey the idea, and might even obscure it somewhat. Nevertheless, it's probably a useful exercise to write it out more formally.
